How to solve $\sqrt{x} + \sqrt{x-\sqrt{1-x}} = 1$ The equation is $$\sqrt{x} + \sqrt{x-\sqrt{1-x}} = 1$$
I cannot find any way to simplify this, other than squaring repeatedly, which is, btw, again not simplifying. Also tried substitutions like $1-x = t^2$ and trigonometric substitutions. None seem to work.
By trial and error, I find one root to be $\dfrac{16}{25}$. Is there a general way to treat this equation?
EDIT
The domain of definition of this function seems to be quite small, some subset of $(0,1)$. Also we know that there is only one root as the function is strictly increasing.
 A: $$\sqrt{x-\sqrt{1-x}}=1-\sqrt{x}$$Squaring we get $$x-\sqrt{1-x}=1-2\sqrt{x}+x\\2\sqrt{x}-\sqrt{1-x}=1$$ Again squaring $$4x-4\sqrt{x(1-x)}+1-x=1\\3x=4\sqrt{x(1-x)}$$And again $$9x^2=16x(1-x)\\9x^2=16x-16x^2\\25x^2-16x=0$$
We can see that $x=0$ is not a root since $\sqrt{x-\sqrt{1-x}}$ is not defined for $x=0$ and checking we see that $x=\frac{16}{25}$ is indeed a root (as you guessed).
A: Following the suggestion in comments, multiply both sides by:
$$\sqrt{x} - \sqrt{x-\sqrt{1-x}}$$
and you get $$\sqrt{1-x}=\sqrt{x}- \sqrt{x-\sqrt{1-x}}$$
Adding to the original equation, and you get:
$$2\sqrt{x}=1+\sqrt{1-x}$$
That's a little easier to solve.
A: Repeated squaring gets me to a quadratic
$$\sqrt{x-\sqrt{1-x}}=1-\sqrt x\\
x-\sqrt{1-x}=1+x-2\sqrt x\\
2\sqrt x-\sqrt{1-x}=1\\
4x+1-x-4\sqrt{x(1-x)}=1\\
9x^2=16x(1-x)\\
x=\frac {16}{25},0$$
and we can check that $0$ does not work.
A: writing your equation in the form
$$\sqrt{x-\sqrt{1-x}}=1-\sqrt{x}$$
squaring we obtain
$$\sqrt{1-x}=2\sqrt{x}-1$$
squaring again
$$4\sqrt{x}=5x$$
and we get
$$25x^2-16x=0$$
thus $$x=0$$ or $$x=\frac{16}{25}$$
A: $$\sqrt{x} + \sqrt{x-\sqrt{1-x}} = 1$$
$$\sqrt{x-\sqrt{1-x}} = 1 -\sqrt{x}$$
$$x-\sqrt{1-x} = (1 -\sqrt{x})^2$$
$$x-\sqrt{1-x} = x- 2\sqrt{x} +1$$
$$\sqrt{1-x} = 2\sqrt{x} +1$$
$$1-x = (2\sqrt{x} +1)^2$$
$$1-x = 1 + 4 \sqrt{x} + 4 x$$
$$- 5x= 4 \sqrt{x} $$
$$- \frac{5}{4}x= \sqrt{x} $$
$$\left(- \frac{5}{4}x\right)^2= x $$
$$\frac{25}{16}x^2= x $$
$$\frac{25}{16}= \frac{1}{x} $$
$$\frac{16}{25}= x $$
